sin(α−3π)+cos(π−α) |
sin(−α)−cos(π+α) |
=
sin(−4π+π+α)−cosα |
−sinα+cosα |
=
sin(π+α)−cosα |
−sinα+cosα |
−sinα−cosα |
−sinα+cosα |
=
sinα+cosα |
sinα−cosα |
tanα+1 |
tanα−1 |
又tan(5π+α)=m,∴tan(π+α)=tanα=m,∴原式=
tanα+1 |
tanα−1 |
m+1 |
m−1 |
故答案为:
m+1 |
m−1 |
sin(α−3π)+cos(π−α) |
sin(−α)−cos(π+α) |
sin(α−3π)+cos(π−α) |
sin(−α)−cos(π+α) |
sin(−4π+π+α)−cosα |
−sinα+cosα |
sin(π+α)−cosα |
−sinα+cosα |
−sinα−cosα |
−sinα+cosα |
sinα+cosα |
sinα−cosα |
tanα+1 |
tanα−1 |
tanα+1 |
tanα−1 |
m+1 |
m−1 |
m+1 |
m−1 |